We know that there are compact manifolds with diffeomorphic interiors but their boundaries are not homeomorphic (see the question Manifolds with homeomorphic interiors).

My question is about a very special case:

Assume $D$ is a bounded open set in $R^n$ with smooth boundary. If $D$ is diffeomorphic to the unit ball, is $\bar D$, the closure of $D$, diffeomorphic to the closed unit ball?


2 Answers 2


The case $n=4$ is open as far as I know.

The case $n=3$ follows since $\mathbb R^3$ is irreducible, so it contains no fake 3-disk, i.e. $\bar D$ must be the standard disk.

The case $n=5$ is equivalent to the smooth $4$-dimensional Poincare conjecture (which is still open). Here is why:

  1. Any homotopy $4$-sphere embeds smoothly into $\mathbb R^5$ (Sketch: homology $4$-sphere bounds a contractible smooth manifold $C$ [Kervaire, "Smooth homology spheres and their fundamental groups", Theorem 3]. In our case $\partial C$ is simply-connected, so attaching a collar on the boundary one gets a contractible $5$-manifold that is simply-connected at infinity, and hence it is diffeomorphic to $\mathbb R^5$ by a result of Stallings).

  2. Any embedding of the standard $4$-sphere into $\mathbb R^5$ bounds a standard disk, see [Smale, "Differentiable and Combinatorial Structures on Manifolds", Corollary 1.3]. What Smale actually states is that any embedded $S^{n-1}$ in $\mathbb R^n$ bounds a standard disk unless $n=4$ or $7$. This was before he proved the h-cobodorsm theorem hence he excludes $7$.

Finally, as mentioned in comments if $n>5$, then $\bar D$ is diffeomorphic to the standard disk by the h-cobordism theorem (sketch: since by assumption $D$ is simply-connected at infinity, $\partial D$ is a homotopy sphere and $\bar D$ is a contractible smooth manifold, so removing a small ball in its interior results in h-cobordism between then standard sphere and the embedded one. Proving that this is an h-cobordism involves standard excision considerations in homology).

  • 2
    $\begingroup$ Thank you for your answer and reference. The case $n=3$ is a classical 3-dimensional topology fact, that doesn't require the Poincaré conjecture (that is, $S^3$ is irreducible), see for ins Hatcher, Notes on Basic 3-Manifold Topology, at the very beginning. $\endgroup$ Feb 23, 2014 at 15:15
  • $\begingroup$ @DanieleZuddas: Oh, indeed, I was not thinking clearly. I will edit. $\endgroup$ Feb 23, 2014 at 15:18
  • $\begingroup$ Can we relax the condition that do not require D us embedded in $R^n$? $\endgroup$
    – J. GE
    Feb 23, 2014 at 15:40
  • $\begingroup$ @J.GE at least, you should request that $D$ is embedded outside of a compact subset of it, and that no points of the critical subset overlap with its embedded complement. $\endgroup$ Feb 23, 2014 at 16:42
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    $\begingroup$ @J.GE: having an embedding is not important: compact smooth $n$-manifold whose interior is an open disk is diffeomorphic to $D^n$, if $n\neq 4,5$. $\endgroup$ Feb 23, 2014 at 20:29

First, your assumption imply that $\bar D$ is a compact smooth manifold with boundary a topological sphere (because is a simply connected homology sphere). So, $\bar D$ is a topological $n$-ball, by the generalized Schoenflies theorem. Now, this reduces to the smooth Schoenflies problem (still open in $\Bbb R^4$). Also, I don't know whether an exotic $(n-1)$-sphere can be smoothly embedded in $\Bbb R^n$.

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    $\begingroup$ Exotic $n$-spheres with $n>4$ cannot be smoothly embedded into $\mathbb R^{n+1}$ by the h-cobordism theorem (the sphere bounds a contractible smooth manifold, so removing a small ball in its interior results in h-cobordism between then standard sphere and the embedded one). I think the same result is true if $n=4$ but the proof must be harder, and I cannot remember it at the moment. $\endgroup$ Feb 23, 2014 at 4:48
  • $\begingroup$ In the last sentence by a "result" I meant that a homotopy $4$-sphere that embeds into $\mathbb R^5$ bounds the standard ball. $\endgroup$ Feb 23, 2014 at 4:57
  • $\begingroup$ @IgorBelegradek thank you for pointing out this... Actually the h-cobordism theorem doesn't hold in this dimension, I think. $\endgroup$ Feb 23, 2014 at 5:27
  • $\begingroup$ What I mean is that (if memory serves) a different argument takes care of smooth Schoenflies in $\mathbb R^5$. $\endgroup$ Feb 23, 2014 at 5:31
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    $\begingroup$ @Misha: by assumption $D$ is a disk, so it is simply-connected at infinity. $\endgroup$ Feb 23, 2014 at 5:48

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